Concept to extract heat energy from objects

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Discussion Overview

The discussion revolves around the concept of extracting heat energy from objects, particularly focusing on the thermal energy contained in materials at room temperature compared to absolute zero. Participants explore the calculations related to heat energy and the practical methods for energy extraction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant calculates the heat energy of 1 kg of glass at room temperature using the equation H = Cp x M x ΔT, arriving at approximately 263550 joules.
  • Another participant notes that while the equation is valid, it is preferable to work in Kelvin, pointing out that the temperature change is equivalent in Celsius and Kelvin.
  • Some participants emphasize that the object is already at room temperature, implying that it has absorbed heat energy from its surroundings.
  • There is a discussion about the efficiency of extracting heat energy, with suggestions that placing the object in a large temperature differential could facilitate energy extraction, likening it to the operation of a fridge.
  • One participant mentions that heat pumps can extract heat from the environment, even at low temperatures, but they require energy to operate, with a typical Coefficient of Performance (COP) ranging from 2 to 6.

Areas of Agreement / Disagreement

Participants express differing views on the methods of extracting heat energy and the implications of the calculations presented. There is no consensus on the most effective approach to energy extraction or the practical applications of the discussed concepts.

Contextual Notes

Some calculations and assumptions regarding specific heat capacity and energy extraction methods remain unverified, and the discussion does not resolve the efficiency of different energy extraction techniques.

Dray1480
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I was just thinking about absolute zero temprature roughly -250 c degress
Lets say an object has a room temprature of 20 degrees celsius
Doesnt it mean i has a lot of thermal emegy
So i used the equation

H= Cp x M x ΔT

H = Heat energy (in Joules)
m = mass (in kilograms)
delta T = change in temperature (in degrees Celsius)
Cp = Specific heat (in J/kg x degrees C)

If the material is glass with mass of 1 kg it would have 263550 joules of energy??

Im just a as student I am completely wrong just say so
 
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Q = M . C . dt (How do you get the triangle symbol) is how I write the equation, but they are the same thing.

You should work in kelvin really, but a change of 1 in either unit is equal so that's no big deal here.

Anyway, 0k is ~-273 degrees celsius.

What your equation says, is that it takes 263550 joules of energy to heat 1 kg of glass from 0k to 293k.

p.s I haven't actually checked your numbers, but assuming they work out.
 
Last edited:
yes but the object is already heated to room temperature from its surroundings
 
Yes, your object at room temperature has 263550 Joules more heat energy per kilogram than the same object at absolute 0. But so what? You titled this "Concept to extract heat energy from objects" and you said nothing about extracting the energy
 
Dray1480 said:
yes but the object is already heated to room temperature from its surroundings

If what you're looking for is an efficient way to 'extract' that energy from an object, you simply need to place the object in a large temperature differential with good conductance.

It's basically a fridge.
 
Dray1480 said:
If the material is glass with mass of 1 kg it would have 263550 joules of energy??

Im just a as student I am completely wrong just say so

Specific Heat capacity of glass is 840 Joules/kg/k so yes 1kg at room temperature 293K contains about 250,000 Joules.

It's not too hard to extract some of it. As lntz said you could put it in a fridge or use another type of heat pump.

Heat pumps for houses work by extracting heat from the outside air or ground and moving it indoors. They can even operate when the outside temperature is below freezing. The problem is that even the best heat pumps need some energy to run. The COP (Coefficient of Performance) for a typical heat pump for houses is typically in the range 2 to 6. In other words they use 1 unit of energy to move between 2 and 6 units of energy from outside to inside.
 

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